Question

The mean weight of loads of rock is 51.0 tons with a standard deviation of 10.0 tons. If 36 loads are chosen at random for a weight check, find the probability that the mean weight of those loads is less than 50.3 tons. Assume that the variable is normally distributed.

Answer 1

0.3372

Explanation:

Mean weight(μ) = 51.0 tons

Standard deviation (σ) = 10.0 tons

n = 36

Pr(x<50.3) = ???

Using normal distribution,

Z = (X - μ)/ σ/√n

When X = 50.3

Z = (50.3 - 51.0) / 10/√36

Z = -0.7/ 10/6

Z = -0.42

From the normal distribution table, 0.42 = 0.1628

Φ(z) = 0.1628

Recall that when Z is negative, Pr(x<a) = 0.5 - Φ(z)

Pr( x < 50.3) = 0.5 - 0.1628

= 0.3372